International Journal of Computer Applications (0975 – 8887) Volume 153 – No5, November 2016 A Mathematical Model for Solving Four Point Test Cross in Genetics Jugal Gogoi Tazid Ali Department of Mathematics Department of Mathematics Dibrugarh University Dibrugarh University Dibrugarh, Assam, India Dibrugarh, Assam, India ABSTRACT same homolog are transmitted together (parental types) into During meiosis cell division, chromosomes are replicated and gametes more often than not during meiosis [7]. Genes may form a tetrad. While the chromosomes are lined up close to assort independently either because they are on different one another, they have the opportunity to interact with one chromosomes or because they are far apart on the same another, get tangled, and recombine to form new chromosome. combinations of alleles. Four point cross refers to using 4 points (genes) to determine the order and distance between the 2.2.Recombination Frequency genes. This test cross is also the method of choice in Recombination frequency is a measure of genetic linkage [7], determining linkage in organisms with many genetic markers. [8] and is used in the creation of a genetic linkage map. In this paper an attempt is made to solve the four point test Recombination frequency (denoted by θ) is the frequency cross problem. with which a single chromosomal crossover will take place between two genes during meiosis. A centimorgan (cM) is a Keywords unit that describes a recombination frequency of 1%. In this Gene, Meiosis, crossing-over, linkage, recombination, three way we can measure the genetic distance between two loci, point test cross, four point test cross. based upon their recombination frequency. Double crossovers (or even numbers of crossovers) would turn into no 1. INTRODUCTION recombination. In this case we cannot tell if crossovers took The principle of independent assortment [1] in genetics place. If the loci we're analyzing are very close (less than 7 assumes that different genes followed in patterns of cM) a double crossover is very rare. When distances become inheritance analysis are on different chromosomes. Mendel higher, the likelihood of a double crossover increases. During studied seven traits [2] in peas: each trait is now known to be meiosis, chromosomes assort randomly into gametes, such located on separate chromosomes. (Mendel’s work was that the segregation of alleles of one gene is independent of rediscovered by Bateson [3] when his research gave progeny alleles of another gene. This is stated in Mendel's Second Law ratios that Bateson could not interpret. While Mendel’s work [8] and is known as the law of independent assortment. The confirmed independent assortment ratios, it was the work of law of independent assortment always holds true for genes Morgan that finally provided an explanation for deviation that are located on different chromosomes, but for genes that from dihybrid ratios). By solving a four point cross we can are on the same chromosome, it does not always hold true. determine two important things: 2.3.Recombination Mapping 1. Order of the genes on a chromosome. Recombination frequencies may be used to map the position of genes (loci) on linear linkage groups. The order of genes 2. The distance (in map units) between each pair of and the relative distances between them in a linkage group genes. corresponds to their order and relative distances on a A successful finding of map distance (i.e., distance between chromosome. Recombination can be used to map genes. two genes) can also help for findings the probability of Morgan hypothesized that the further apart two genes are, the recombination [4]. more likely they are to have a crossover between them. Sturtevant [10] worked out method to map genes on 2. DEFINITIONS AND ASSUMPTIONS chromosomes. Genes far enough apart on same chromosome 2.1.Gene can show recombination frequencies of 50%. Without A gene is a hereditary factor [5] that had two or more alleles addition all information, one cannot distinguish these cases which determined the difference between two or more from loci on independently segregating chromosomes. Map alternative phenotypes. Different genes controlled different function is a mathematical formula that relates the observed number of crossovers to the real number, which is a function aspects of phenotype. Different genes could be separated by of the physical distance [11]. It assumes a simple model in recombination. So gene is the unit of recombination also. which crossovers are distributed randomly on the These units usually agreed with each other, until genetic chromosome. The maximum recombination frequency is analysis was extended to bacteria and viruses in which rare 50%: as distance increases, crossovers between any two genotypes can be selected and detected. Mutations [6] within markers increases, but some of these are double and other a gene, even in adjacent base pairs, can be separated by even-numbered crossovers. The recombination frequencies of recombination. Two different genes identified by pairs of genes indicate how often 2 genes are transmitted recombination or mutation may control the same phenotype. together. For linked genes, the frequency is less than 50% . Genes on the same chromosome are physically connected or The greater the distance between linked genes, the higher the linked. Gene pairs that are close together on the same recombination frequency (푟. 푓). Recombination frequencies chromosome are genetically linked because the alleles on the 45 International Journal of Computer Applications (0975 – 8887) Volume 153 – No5, November 2016 become more inaccurate as the distance between genes vg+ b+ pr 9 increases. The relationship between relative recombination frequency vg+ b pr+ 131 and distance is used to create genetic maps. The greater the density of genes on the map, and the smaller the distances between the genes results more accurate the map. vg+ b+ pr+ 1654 2.4.Existing Method for solving Three-Point vg b pr+ 13 Test Crosses Step 1. In a three point cross first rewrite the classes of progeny (data) assigning genotypes to each trait. vg b+ pr+ 241 Step 2. The offspring are generated by a heterozygous individual. Therefore, all classes (recombinant, parental, etc.) Where, will occur as reciprocal pairs of progeny. These reciprocal All three alleles that are on a single chromosome are pairs will be both genetic reciprocals and numerically written together. equivalent. SCO- single crossover Step 3. Designate the different gametes or offspring as noncrossover (parental), single crossover or double crossover. DCO- double crossover The noncrossover classes are those classes of progeny who have one of the intact, non-recombinant homologs from the • First, the offspring need to be arranged into parent. The noncrossover classes will be represented by the reciprocal pairs of no crossovers, SCO, and DCO as greatest numbers of offspring. The double crossover classes follows: will be represented by the smallest numbers of offspring. vg b pr 1779 Sometimes one or both double crossover classes are missing because they are rare. vg+ b+ pr+ 1654 Step 4. By examining the pattern of data seen in a problem, vg+ b pr 252 one can often start solving the problem with a basic understanding of the linkage relationships of the genes. Some vg b+ pr+ 241 of the more common patterns of data are: vg+ b pr+ 131 • 3 linked genes give 8 classes of data that occur as 4 vg b+ pr 118 reciprocal pairs genetically and numerically; vg b pr+ 13 • 3 unlinked genes gives 8 classes of data that occur as 4 genetically reciprocal pairs, but all classes are vg+ b+ pr 9 seen in a 1:1:1:1:1:1:1:1 ratio; • Because there is a higher probability of having no Step 5. Begin the process of mapping the genes by ordering crossovers, the two genotypes with the highest the genes. To figure out which gene is in the middle of a number of offspring are NCO (no crossover). There group of three genes, choose one of the double crossover are two sets of SCO, and the two with the lowest classes. Compare it to the most similar parental class of offspring will be DCO. progeny where two of the three genes will have the same combination of alleles. The gene that differs is the gene in the Next we need to determine which order of alleles on the middle. chromosome will result in the DCO. Test problem 1 • Here the no crossovers are (i.e., highest number of To draw a map of these 3 genes (푣푔, 푏, and 푝푟) showing the offspring) distances between all pairs of genes of the following: vg b pr Drosophila autosomal genes: vg+ b+ pr+ 풗품 = vestigial wings; 풗품+ = normal • And the DCO are (i.e., lowest number of offspring) 풃 = black body; 풃+ = normal body vg b pr+ 풑풓 = purple eyes; 풑풓+ = normal eyes. vg+ b+ pr GAMETE OBSERVED FREQUENCY • In order for this to be a double crossover, the 풑풓 allele must be in the center, i.e., they are as follows: vg b pr 1779 vg pr b vg b+ pr 118 3. FOUR-POINT CROSSES A four point test cross can reveal map order [Figure 1]. A successful approach is to analyze results by calculating vg+ b pr 252 recombination frequencies per gene pair. If the parental cross is not known, then the parental types are those progeny classes that are greater than 50% of the total. The occurrence 46 International Journal of Computer Applications (0975 – 8887) Volume 153 – No5, November 2016 of one cross-over event in one region of a chromosome Gl lg a Df 22 decreases the likelihood of another cross-over in an adjacent region. This is known as interference [12], and can be gl Lg A N 41 calculated based on observed and expected recombination frequencies.